Question: The equation of a circle $C$ is $x^2+y^2-4x+18y+69 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Explanation: To find the equation in standard form, complete the square. $(x^2-4x) + (y^2+18y) = -69$ $(x^2-4x+4) + (y^2+18y+81) = -69 + 4 + 81$ $(x-2)^{2} + (y+9)^{2} = 16 = 4^2$ Thus, $(h, k) = (2, -9)$ and $r = 4$.